Calculate Elevator's Acceleration with Person on Bathroom Scale

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To calculate the elevator's acceleration, consider the forces acting on the person standing on the scale. Initially, the person's weight is determined by the equation w=mg, where g is 32.2 ft/sec². When the elevator accelerates, the scale reads only 0.67 of the person's weight, indicating a change in the net force experienced. This change can be analyzed using Newton's second law, where the resultant force equals mass times acceleration. The discussion emphasizes the importance of correctly applying these principles to find the elevator's acceleration.
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A person stands on a bathroom scale in a motionless elevator. When the elevator begins to move, the scale briefly reads only 0.67 of the person's regular weight. Calculate the acceleration of the elevator.

how do i start this problem? help please!
 
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do you have any equations of motion you are suppose to use?
 
<tex>w=mg<txt>
when he's not accelerating his weight is something and g=32.2 ft/sec2. When the elevator accelerates his weight drops to .67 of what it was. (his mass hasn't changed)
 
<tex> w=mg<tex> I really wish I could make LATEX work.
 
Paulanddiw said:
<tex> w=mg<tex> I really wish I could make LATEX work.

Use [ and ] instead of < and >
 
The person is experiencing two forces, his weight and the force from the scale (the reading on the scale is caused by the action/reaction force of this). The vector sum of this two forces gives the resultant "elevator acccelerating force" on the person according to N2.
 
hotvette said:
Use [ and ] instead of < and >

And don't forget the / before the ending "tex":

[ tex ] e^{-x^2} [ /tex ]

gives e^{-x^2}.

The "Math Message Board" uses < > instead of [ ] and I am forever getting them mixed up!
 
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